Question: If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$?
Answer: We solve for $y$ in terms of $x$: \[y = \frac{100 - 5x}{3}.\] Then we express $xy$ in terms of $x$: \[xy = x\frac{100 - 5x}{3} = \frac{100x - 5x^2}{3} = -\frac{5}{3}x^2 + \frac{100}{3}x.\] The graph of this expression is a parabola facing downwards.  The greatest possible value of $xy$ occurs at the vertex of this parabola, which occurs when $x = \frac{-100/3}{2\cdot -5/3} = 10$.  Then,  \[xy = 10\cdot \frac{50}{3} = \frac{500}{3}.\]   However, this is not an integer.  So, we test the two closest integer values of $x$: $x=9$ and $x=11$, to see if either of these yield integer values for $y$.

When $x=9$, $y=\frac{55}{3}$, which is not an integer.  When $x=11$, $y=\frac{45}{3}=15$, which is an integer.  In this case, \[xy = 11\cdot 15 = \boxed{165}.\]